for \({r}_{e1}\), we consider the centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}+{r}_{e2})}^2}\)
for \({r}_{e2}\),
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}+{r}_{e2})}^2}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac {2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } } }{ \cfrac {2 }{ { r }^{ 2 }_{ e2 } } -\cfrac { 1 }{ { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } } } \)
\( \cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { { 2{ ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\(\cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { { 2{ ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\( x=\cfrac { 2(x+1)^{ 2 }-x^2 }{ 2(x+1)^{ 2 }-1 } \)
\(x\) has three solutions. The only positive solution is,
\(x\) = 1, \({r}_{e1}={r}_{e2}\)
this suggest that this configuration is possible.
And for the next extreme case, without lost of generality, \({r}_{e1}\gt{r}_{e2}\), the diagram on the right
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}-{r}_{e2})}^2}\)
and
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}+\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}-{r}_{e2})}^2}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac { 2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } } }{ \cfrac { 2 }{ { r }^{ 2 }_{ e2 } } +\cfrac { 1 }{ { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } } } \)
\( \cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }+{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\(\cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }+{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\( x=\cfrac {2 (x-1)^{ 2 }-x^2 }{2 (x-1)^{ 2 }+1 } \)
\(x\) also has three solution, a complex pair, and a positive solution,
\(x = 0.3684\), \({r}_{e1}=0.3684*{r}_{e2}\)
Since we have assumed that \({r}_{e1}\gt{r}_{e2}\), this solution is not admissible. This configuration is not possible. This might suggest that the electrons will always remain on the opposite side of the positive charge.
We will consider the limiting case of,
For \({r}_{e1}\), again we look at the centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}-\cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac { 2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ \sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) } }{ \cfrac { 2 }{ { r }^{ 2 }_{ e2 } } -\cfrac { 1 }{ { \sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) } } } \)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e1 } } }{ { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\( \cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e1 } } }{ { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\(x=\cfrac { { 2\sqrt { 2 } ({ x }^{ 2 }+1^{ 2 }) }-{ { x }^{ 2 } } }{ { 2\sqrt { 2 } ({ x }^{ 2 }+1^{ 2 }) }-1 }\)
\(x\) also has three solution, a complex pair, and a positive solution,
\(x =1\), \({r}_{e1} = {r}_{e2}\)
This is possible. In this case there is a tangential force component equal to
\({F}_{t}= \cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
but the speed of the electron will not increase further. Speed of the electron is at light speed \(c\). As the electron move relative to one another, the centripetal force changes and the orbital radius changes. If we now consider a small displacement \(\Delta a\) towards the other electron,
We see that the electrons will experience a force that drive them further apart. And the electrons will remain on opposite sides of the positive charge.
